Let $Q_1$ and $Q_2 $ be circumferences with centers $ O_1$ and $O_2$ respectively tangent outwardly. Let $A$ and $B$ be points on $Q_1$ and $Q_2$, respectively, such that the line $AB ^ {\leftrightarrow}$ is a common tangent external to $Q_1$ and $Q_2$. Let $C$ and $D $ be points in the semiplane determined by $AB ^ {\leftrightarrow}$ that does not contain $O_1$ and $O_2$ such that ABCD is a square. If O is the center of this square, determine the possible values of angle $O_1OO_2$
Maybe it could involve the motto: Lemma. Let two nonintersecting circles $\odot O_1$ and $\odot O_2$ be given. Let an internal common tangent touch the circles at $S, S'$ and intersect the two external common tangents at $Q, Q'$. Then $SQ = S'Q'$.
Let $r_1$ and $r_2$ be the radii of the two circles and $a$ be the side length of the square. We have,
$$a^2 = (r_1+r_2)^2-(r_1-r_2)^2 = 4r_1r_2$$
Apply the cosine rule to the triangles $OAO_1$ and $OBO_2$,
$$OO_1^2 = r_1^2 + \frac12a^2-2r_1\frac{a}{\sqrt2}\cos135=r_1^2+2r_1r_2+2r_1\sqrt{r_1r_2}$$ $$OO_2^2 = r_2^2 + \frac12a^2-2r_2\frac{a}{\sqrt2}\cos135=r_2^2+2r_1r_2+2r_2\sqrt{r_1r_2}$$
Now, apply the cosine rule to the triangle $OO_1O_2$ to express the angle $\theta = \angle O_1OO_2$,
$$\cos\theta = \frac {OO_1^2+OO_2^2-(r_1+r_2)^2}{2\>OO_1\cdot OO_2}$$ $$=\frac{1+\sqrt{\frac{r_1}{r_2}}+\sqrt{\frac{r_2}{r_1}}}{\sqrt{\left(1+2\frac{r_1}{r_2}+2\sqrt{\frac{r_1}{r_2}}\right)\left(1+2\frac{r_1}{r_2}+2\sqrt{\frac{r_1}{r_2}}\right)}}$$ $$=\frac{1+\sqrt r+\frac{1}{\sqrt r}}{\sqrt{ \left( 1+2r+2\sqrt{r} \right) \left( 1+\frac 2r +\frac{2}{\sqrt r} \right) }}$$
where $r=\frac{r_1}{r_2}$, the ratio of the two circles. Now, examine the two limits.
1) If $r=1$, $\cos\theta = \frac35$;
2) If $r=\infty$, $\cos\theta = \frac{1}{\sqrt{2}}$.
Therefore, the possible values of the angle $\theta = \angle O_1OO_2$ are in the range
$$45^\circ < \theta \le \cos^{-1}\frac35$$